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keche0b

Answered question

2021-12-27

Find the general solution of the following differential equation.
$y^{″} + 2 y^{'} + 5 y = 0$

Answer & Explanation

Hattie Schaeffer

Beginner2021-12-28Added 37 answers

y 2 y^{'} + 5 y = 0

Let,

y = A e^{m x}

be the trial solution at (2)
Hence, the auxiliary equation is

m^{2} + 2 m + 5 = 0

Using quadratic formula

m = \frac{- 2 \pm \sqrt{4 - 20}}{2}

\Rightarrow m = \frac{- 2 \pm \sqrt{- 16}}{2}

\Rightarrow m = \frac{- 2}{2} \pm \frac{4 i}{2} (∴ i = \sqrt{- 1})

m = - 1 \pm 2 i

Therefore, the general solution is

y = e^{- x} (c_{1} \cos 2 x + c_{2} \sin 2 x)

Alex Sheppard

Beginner2021-12-29Added 36 answers

y 2 y^{'} + 5 y = 0

k^{2} + 2 k + 5 = 0

D = 4 - 20 = - 16

k_{1} = - 1 + 2 i

k_{2} = - 1 - 2 i

y = (C_{1} \sin 2 x + C_{2} \cos 2 x) e^{- x}

karton

Expert2022-01-09Added 613 answers

$\begin{matrix} K^{2} - 2 k + 5 = 0 \\ D = \sqrt{4 - 20} = \sqrt{- 16} \\ i^{2} = - 1, D = {\sqrt{16 i}}^{2} = \pm - 4 i \\ k_{1} = (2 - 4 i) / 2 = 1 - 2 i \\ k_{2} = (2 + 4 i) / 2 = 1 + 2 i \\ Y = c_{1} e^{1 - 2 i} x + c 2 e^{1 + 2 i} x \end{matrix}$

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